waecmaths question:
Simplify $\sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}$
Option A:
16
Option B:
8
Option C:
4
Option D:
1
waecmaths solution:
$\begin{align} & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{\frac{{{({{2}^{3}})}^{2}}\times {{2}^{2(n+1)}}}{{{2}^{2n}}\times {{2}^{4}}}}=\sqrt{\frac{{{2}^{6}}\times {{2}^{2n+2}}}{{{2}^{2n}}\times {{2}^{4}}}} \\ & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{\frac{{{2}^{8+2n}}}{{{2}^{2n+4}}}}=\sqrt{{{2}^{8+2n-2n-4}}} \\ & \sqrt{\frac{{{8}^{2}}\times {{4}^{n+1}}}{{{2}^{2n}}\times 16}}=\sqrt{{{2}^{4}}}={{({{2}^{4}})}^{\tfrac{1}{2}}}={{2}^{2}}=4 \\\end{align}$
maths year:
maths topics:
